React Native - 单选单选

IT技术 javascript android reactjs react-native mobile
2021-04-28 05:47:54

我一直在寻找如何在Flatlist 中进行单一选择,但我找不到任何,在我的代码中,我试图对我的Flatlist内的每个单元格进行单一选择

示例:我选择第 1 号单元格,然后将选择第 1 号单元格。如果我需要选择2号,1号和2号都会被选择。

我的代码中发生的事情是当我选择 1 号时,它会选择所有单元格。

export default class Dishes extends Component {
    constructor(props) {
        super (props)
        this.state = {
            data: [],
            id: [],
            price: [],
            count: 0,
            SplOrder: '',
            tbl: this.props.navigation.state.params.tbl,
            orderDet: this.props.navigation.state.params.orderDet,
            DineIn: this.props.navigation.state.params.DineIn,
            TakeOut: this.props.navigation.state.params.TakeOut,
        }
    }

/********************EDIT*************************
_incrementCount() {
    this.setState({ count: this.state.count + 1 });
}
_decreaseCount() {
    this.setState({ count: this.state.count - 1 });
}
changeTextHandler() {
    this.setState({ ['count'+index]: text });
};
*/

    _renderItem = ({ item, index }) => {
        return (
            <View>
                <View>
                    <View>
                        <Text>Name: { item.menu_desc }</Text>
                            <View}>
                                <Text>Price: ₱{ item.menu_price }</Text>
                                <Text>Status: { item.menu_status }</Text>
                            </View>
                        <TextInput
                            placeholder = "Insert Special Request"
                            onChangeText = { (text) => this.setState({ SplOrder: text }) }
                            value = { this.state.SplOrder }
                        />
                    </View>
                    <View>
                        <TouchableOpacity
                            onPress = {() => this._incrementCount()}>
                            <Text> + </Text>
                        </TouchableOpacity>
                        <TextInput
                            onChangeText={this.changeTextHandler}
                            value={this.state['count'+index].toString()}    // Not working
                            placeholder="0"/>
                        <TouchableOpacity
                            onPress = {() => this._decreaseCount()}>
                            <Text> - </Text>
                        </TouchableOpacity>
                    </View>
                </View>
            </View>
        )
    }

    render() {
        return (
            <View>
                <View>
                    <Text>Table No: { this.state.tbl }</Text>
                    <Text>Order No: { this.state.orderDet }</Text>
                    <Text>{ this.state.DineIn }{ this.state.TakeOut }</Text>
                </View>

                <FlatList
                data = {this.state.data}
                keyExtractor={(item, index) => index.toString()}
                extraData={this.state}
                renderItem = {this._renderItem}
                />
                <View>
                    <TouchableOpacity
                    onPress = {() => this.submit()}>
                        <Text>Send Order</Text>
                    </TouchableOpacity>
                </View>
            </View>
        )
    }
}
1个回答

TextInputs瞄准相同的状态SplOrder的话,如果你改变它在任何TextInput其他人将显示相同。我看到的解决方案是为您创建的每个项目设置一个状态。你应该做这个:

                    <TextInput
                        placeholder = "Insert Special Request"
                        onChangeText = { (text) => this.setState({ ['SplOrder'+index]: text }) }
                        value = { this.state['SplOrder'+index] }
                    />

我们在评论中讨论的第二个问题应该通过将 index 参数传递给 incrementCount 函数来解决。

嗨,现在在方法 _incrementCount() 中,您必须传递索引并使用它的索引递增每个计数,就像您在值中拥有的那样。所以你可以这样做

<TouchableOpacity
    onPress = {() => this._incrementCount(index)}>
    <Text> + </Text>
</TouchableOpacity>

然后更改您的 _incrementCount 方法,添加一个参数并执行如下操作:

_incrementCount(index) {
    this.setState({ ['count'+index]: this.state['count'+index] + 1 });
}